Trigonometric Identities

You have studied various algebraic identities so far. Similarly, we have various trigonometric identities, which are true for all variables.

Now, let us prove these identities.

Let us take a standard circle with radius r such that it intersects the X-axis at point A. Also, let the initial arm OA is rotated in anti-clockwise direction by an angle ?.

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_18_17_35_02/5.png

In the figure, the terminal arm intersects the circle at point P (x, y) where x, y ≠ 0 and OP = r. By the definition of trigonometric ratios, we have

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_18_17_35_02/mathmlequation5750724437484245635.png

Now, OP is a distance between origin O (0, 0) and point P (x, y) which can be obtained by distance formula as follows:

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_18_17_35_02/mathmlequation8813183984335007542.png

(1) On dividing both sides of the equation (i) by r2, we get

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_18_17_56_11/mathmlequation4736426815110703942.png

(2) On dividing both sides of the equation (i) by x2 (x ≠ 0), we get

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_18_17_56_11/mathmlequation6517315116028025759.png

(3) On dividing both sides of the equation (i) by y2 (y ≠ 0), we get

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_18_17_56_11/mathmlequation7636672636949875878.png

Corollary:

(i)  When x = 0 then we have

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_18_17_35_02/mathmlequation7058086621640685920.png

In this case, the identities sin2θ + cos2θ = 1 and 1 + cot2θ = cosec2θ exist but the identity 1 + tan2θ = sec2θ does not exist.

(ii)  When y = 0 then we have

https://img-nm.mnimgs.com/img/study_content/editlive_lp/56/2013_04_18_17_35_02/mathmlequation7043690161630004156.png

In this case, the identities sin2θ + cos2θ = 1 and 1 + tan2θ = sec2θ exist but the identity 1 + cot2θ = cosec2θ does not exist.

Now, we know the basic trigonometric identities, let us see the following video to know how to use these identities.

Example 1: Find the value of the expression (sec227° – tan27°.cot63°).

Solution:

sec227° – tan27°.cot63° = sec227° – tan27°.cot (90° – 27°)

[27° and 63° are complementary angles]

= sec227°– tan27°.tan27° [cot (90° – ?) = tan ?]

= sec227° – tan227°

= 1 + tan227° – tan2 27° [Using the identity 1 + tan2 θ = sec2 θ]

= 1

Thus, the value of the given expression is 1.

Example 2: Write all the trigonometric ratios in terms of sin A.

Solution:

Using the identity

sin2A + cos2A = 1,

we can write, cos2A = 1 – sin2A

Taking square root on both sides,

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_m477c77f5.gif

Now,https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_2e1eb8f4.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_42b18e6f.gif[Using (i)]

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_1f4bf011.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_m4ae9f89b.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_6237c160.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_2e85a310.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_m47943fed.gif [Using (i)]

and,https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_58bdc0f7.gif

The trigonometric ratios in terms of sin A are given by (i), (ii), (iii), (iv), and (v).

Example 3: Simplify the following expression.

[(1 + cot A – cosec A)(1 + tan A + sec A)]

Solution:

(1 + cot A – cosec A)(1 + tan A + sec A)

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/457/953/10.1.8.4.1_ok_SS_html_75ce707c.gif

Thus, the value of the given expression is 2.

Use Of Trigonometric Identities In Proving Relationships Involving Trigonometric Ratios

Can we prove the following relation?

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_37cad9b7.gif

To prove this relation, we will make use of the following identities.

Let us see how to prove this.

We can take any side, L.H.S or R.H.S, of the above relation and prove it equal to the other side.

Now, L.H.S. =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_4ea46bff.gif

= https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_m15f655d2.gif(by rationalising the denominator)

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_ma7aef7.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_7ffc93c0.gif = (Using the identity 1 + cot2θ = cosec2θ)

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_m5313c342.gif

= sec A + tan A

= R.H.S.

Thus,https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_m70ab0b56.gif

In this way, we use the above mentioned trigonometric identities to prove other relations involving trigonometric ratios.

Let us now look at another approach to solve a relation.

In this method, we simplify both L.H.S and R.H.S. and when we reach at a common point, we can say that the L.H.S. and R.H.S. are equal.

Let us try to prove that 2tan2A + tan4A = sec4A − 1.

Taking the L.H.S of the equation,

The L.H.S is given as 2tan2A + tan4A.

= tan2A (2 + tan2A)

= (sec2A − 1) (2 + sec2A − 1) [Using the identity 1 + tan2A = sec2A, we obtain tan2A = sec2A − 1]

= (sec2A − 1) (sec2A + 1)

The R.H.S is given by sec4A − 1.

= (sec2A − 1) (sec2A + 1) [a2 − b2 = (a − b) (a + b)]

Therefore, by solving both L.H.S and R.H.S, we obtain (sec2A − 1) (sec2A + 1).

Thus, we can say that 2tan2A + tan4A = sec4A − 1.

Now let us look at some more examples.

Example 1: Prove that https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_m79d95676.gif.

Solution:

L.H.S =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_m1cb39f1b.gif

[Multiplying and dividing by (1 + sin A)]

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_2597a25.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_542d69e8.gif[Using the identity cos2A + sin2A = 1]

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_m402ef096.gif

= R.H.S

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_2c8dd0c1.gif

Example 2: Prove thathttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_33418777.gif

Solution:

L.H.S = (cosec A − sin A) (sec A − cos A)

= cosec A × sec A − cosec A × cos A − sin A × sec A + sin A × cos A

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_39cb36fb.gif+ sin A × cos A

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_m6280bcfd.gif

And R.H.S is given by

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_2b6f1ed3.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_m576b001f.gif

= sin A × cos A [∵sin2A + cos2A = 1]

Now L.H.S = R.H.S = sin A × cos A

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/181/458/552/10.1.8.4.2_SMT_PIY_KSB_SS_html_2f74dbd6.gif